I would of thought the camera is of irrelevance? The question is about how you interpret 1/10th stops on a light meter..... I was just asking should I read 5/10ths as a third of a stop? My camera is a 5D MKII

Exposure isnt an exact science. Sometimes the 'best' exposure is technically slightly over or under exposed, thats what exposure compensation is for. Digital cameras make this so easy for us. We can just take the shot, and if it isnt exposed correctly, we can look and change the settings slightly.

I think that's as close as you're going to get. Beware that the whole thing is logarithmic, so 0.5 stops from F/8 to F/11 is not F/9.5, but more like F/9.

Yes, and its log base 2. Actually a simple solution to such problems as to do a little maths.

If any thing says: X + yStops (y can also be a fraction)
Then the required aperture is: X * 2^(y/2)

So in this example: 2.8 + 5/10
= 2.8 * 2^(5/20)
= 2.8 * 2^(1/4)
= 2.8 * 1.1892
= 3.3297
~ 3.33
But we do not have 3.33 so go to 3.5, thats the closest you can get. BTW 3.5 is actually 2/3stop down 2.8.

Similarly, For 1.4 + 5/10 gives 1.66 which is not not available so the closest will be 1.8. Again 1.8 is 2/3 stop down 1.4.

You see the problem here is that the aperture quantization is done in thirds not in halfs. But, if you really want your exposure to be spot on. Try nutral density filters. these filters can be combined for all kinds of fractional